Let $\{a_n\}$ be a sequence of strictly increasing positive real numbers diverging to infinity such that $\lim_{n \to \infty} \dfrac {a_n}{a_{n+1}}=1$ , then is it true that the set $\{a_n/a_m : m,n \in \mathbb N , m\ge n\}$ is dense in $[0,1]$ ? I am only able to show that $0$ is a closure point of the given set of ratios . Please help. Thanks in advance


You know that $1$ is a limit point because of $\lim_{n\rightarrow\infty}\frac{a_n}{a_{n+1}}=1$ (and $a_n\not=a_{n+1}$). You also know that $0$ is a limit point because $\lim_{n\rightarrow\infty}\frac{a_1}{a_n}=0$ as $a_n$ diverges to infinity.

Observe that for $y\in[0,1]$, the difference between $y$ and $y(1-\varepsilon)$ is $y\varepsilon$ which is at most $\varepsilon$.

Let $x\in(0,1)$. Fix $\varepsilon>0$. Choose $N$ sufficiently large so that for all $n\geq N$, $\frac{a_n}{a_{n+1}}>1-\varepsilon$ and $\frac{a_n}{a_{n+1}}>x$.

Consider the sequence $\{\frac{a_N}{a_m}:m>N\}$. Observe the following:

  1. the first term in the sequence is greater than $x$
  2. each term in the sequence can be found from the previous term by multiplying by $\frac{a_{m}}{a_{m+1}}$ (each of which is greater than $1-\varepsilon$).
  3. the limit $\lim_{m\rightarrow\infty}\frac{a_N}{a_m}=0$.

Therefore, this sequence includes points in $[0,x]$ and any point in $[0,x]$ is within $\varepsilon$ of a point of the sequence (you've got a sequence decreasing to zero, each step of the sequence is within $\varepsilon$ of the previous step, and it starts before $x$). So, in particular, we can choose the point that is within $\varepsilon$ of $x$. By varying $\varepsilon$, we show that $x$ is a limit point.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy